标签：max clu 节点 scan bzoj uniq 区间 stat 需要
马上就要noi了……可能滚粗已经稳了……但是还是要复习模板啊
LCT: bzoj2049 1A 7min
# include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int M = 5e5 + 10; const int mod = 1e9+7; struct LCT { int ch[M][2], fa[M], sz[M]; bool rev[M]; # define ls ch[x][0] # define rs ch[x][1] inline void up(int x) { if(!x) return ; sz[x] = 1 + sz[ls] + sz[rs]; } inline void pushrev(int x) { if(!x) return ; rev[x] ^= 1; swap(ch[x][0], ch[x][1]); } inline void down(int x) { if(!x) return ; if(rev[x]) { pushrev(ls); pushrev(rs); rev[x] = 0; } } # undef ls # undef rs inline bool isrt(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; } inline void rotate(int x) { int y = fa[x], z = fa[y], ls = ch[y][1] == x, rs = ls^1; if(!isrt(y)) ch[z][ch[z][1] == y] = x; fa[ch[x][rs]] = y; fa[y] = x, fa[x] = z; ch[y][ls] = ch[x][rs]; ch[x][rs] = y; up(y); up(x); } int st[M]; inline void splay(int x) { int stn = 0, tx = x; while(!isrt(tx)) st[++stn] = tx, tx = fa[tx]; st[++stn] = tx; for (int i=stn; i; --i) down(st[i]); while(!isrt(x)) { int y = fa[x], z = fa[y]; if(!isrt(y)) { if((ch[z][0] == y) ^ (ch[y][0] == x)) rotate(x); else rotate(y); } rotate(x); } } inline int access(int x) { int t = 0; for (; x; t=x, x=fa[x]) { splay(x); ch[x][1] = t; up(x); } return t; } inline void makeroot(int x) { access(x); splay(x); pushrev(x); } inline int find(int x) { access(x); splay(x); while(ch[x][0]) x = ch[x][0]; return x; } inline void link(int x, int y) { makeroot(x); fa[x] = y; } inline void cut(int x, int y) { makeroot(x); access(y); splay(y); ch[y][0] = fa[x] = 0; up(y); } }T; int n, Q, x, y; int main() { static char op[23]; cin >> n >> Q; for (int i=1; i<=n; ++i) { T.ch[i][0] = T.ch[i][1] = T.fa[i] = 0; T.sz[i] = 1; T.rev[i] = 0; } while(Q--) { scanf("%s%d%d", op, &x, &y); if(op[0] == ‘C‘) T.link(x, y); else if(op[0] == ‘D‘) T.cut(x, y); else puts(T.find(x) == T.find(y) ? "Yes" : "No"); } return 0; }
dsu on tree: bzoj4756 1A 10min
# include <vector> # include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int N = 1e5 + 10; const int mod = 1e9+7; int n, a[N]; vector<int> ps; int head[N], nxt[N], to[N], tot = 0; inline void add(int u, int v) { ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; } struct BIT { int n, c[N]; # define lb(x) (x & (-x)) inline void set(int _n) { n = _n; memset(c, 0, sizeof c); } inline void edt(int x, int d) { for (; x<=n; x+=lb(x)) c[x] += d; } inline int sum(int x) { int ret = 0; for (; x; x-=lb(x)) ret += c[x]; return ret; } inline int sum(int x, int y) { if(x > y) return 0; return sum(y) - sum(x-1); } # undef lb }T; int sz[N], mx[N]; inline void dfs(int x) { sz[x] = 1; mx[x] = 0; for (int i=head[x]; i; i=nxt[i]) { dfs(to[i]); sz[x] += sz[to[i]]; if(mx[x] == 0 || sz[to[i]] > sz[mx[x]]) mx[x] = to[i]; } } int ans[N], big[N]; inline void count(int x, int p) { T.edt(a[x], p); for (int i=head[x]; i; i=nxt[i]) if(!big[to[i]]) count(to[i], p); } inline void dsu(int x, bool kep = 0) { int son = mx[x]; for (int i=head[x]; i; i=nxt[i]) if(to[i] != son) dsu(to[i], 0); if(son) big[son] = 1, dsu(son, 1); count(x, 1); ans[x] = T.sum(a[x] + 1, n); if(son) big[son] = 0; if(!kep) count(x, -1); } int main() { cin >> n; T.set(n); for (int i=1; i<=n; ++i) { scanf("%d", a+i); ps.push_back(a[i]); } sort(ps.begin(), ps.end()); ps.erase(unique(ps.begin(), ps.end()), ps.end()); for (int i=1; i<=n; ++i) a[i] = lower_bound(ps.begin(), ps.end(), a[i]) - ps.begin() + 1; for (int i=2, par; i<=n; ++i) { scanf("%d", &par); add(par, i); } dfs(1); dsu(1); for (int i=1; i<=n; ++i) printf("%d\n", ans[i]); return 0; }
线段树合并: bzoj4756 2A 13min
warning: 需要注意节点数量为$O(nlogn)$。
# include <vector> # include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int N = 1e5 + 10, SN = 262144 * 20 + 5; const int mod = 1e9+7; int n, a[N]; vector<int> ps; int head[N], nxt[N], to[N], tot = 0; inline void add(int u, int v) { ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; } int rt[N]; struct SMT { int siz; int s[SN], ch[SN][2]; # define ls ch[x][0] # define rs ch[x][1] inline void set() { siz = 0; memset(ch, 0, sizeof ch); memset(s, 0, sizeof s); } inline void up(int x) { if(!x) return; s[x] = s[ls] + s[rs]; } inline void edt(int &x, int l, int r, int ps, int d) { if(!x) x = ++siz; if(l == r) { s[x] = 1; return; } int mid = l+r>>1; if(ps <= mid) edt(ls, l, mid, ps, d); else edt(rs, mid+1, r, ps, d); up(x); } inline int sum(int x, int l, int r, int L, int R) { if(!x || L>R) return 0; if(L <= l && r <= R) return s[x]; int mid = l+r>>1, ret = 0; if(L <= mid) ret += sum(ls, l, mid, L, R); if(R > mid) ret += sum(rs, mid+1, r, L, R); return ret; } inline int merge(int x, int y, int l, int r) { if(!x || !y) return x+y; if(l == r) { s[x] += s[y]; return x; } int mid = l+r>>1; ls = merge(ch[x][0], ch[y][0], l, mid); rs = merge(ch[x][1], ch[y][1], mid+1, r); up(x); return x; } }T; int ans[N]; inline void dfs(int x) { for (int i=head[x]; i; i=nxt[i]) { dfs(to[i]); rt[x] = T.merge(rt[x], rt[to[i]], 1, n); } ans[x] = T.sum(rt[x], 1, n, a[x] + 1, n); } int main() { cin >> n; T.set(); for (int i=1; i<=n; ++i) { scanf("%d", a+i); ps.push_back(a[i]); } sort(ps.begin(), ps.end()); ps.erase(unique(ps.begin(), ps.end()), ps.end()); for (int i=1; i<=n; ++i) a[i] = lower_bound(ps.begin(), ps.end(), a[i]) - ps.begin() + 1; for (int i=2, par; i<=n; ++i) { scanf("%d", &par); add(par, i); } for (int i=1; i<=n; ++i) T.edt(rt[i], 1, n, a[i], 1); dfs(1); for (int i=1; i<=n; ++i) printf("%d\n", ans[i]); return 0; }
线段树：codevs4927 2A 8min
warning: 区间覆盖后线清空tag标记，然后再传下来的tag标记是在cov之后的，所以要先传cov再传tag
# include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> # ifdef WIN32 # define LLD "%I64d" # else # define LLD "%lld" # endif using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int N = 1e5 + 10, SN = 262144 + 5; const int mod = 1e9+7; int n, m; ll a[N]; struct SMT { ll s[SN], mx[SN], mi[SN], tag[SN], cov[SN]; bool hc[SN]; # define ls (x<<1) # define rs (x<<1|1) inline void up(int x) { s[x] = s[ls] + s[rs]; mx[x] = max(mx[ls], mx[rs]); mi[x] = min(mi[ls], mi[rs]); } inline void pushtag(int x, int l, int r, ll tg) { tag[x] += tg; s[x] += tg*(r-l+1); mx[x] += tg, mi[x] += tg; } inline void pushcov(int x, int l, int r, ll tg) { tag[x] = 0; hc[x] = 1; cov[x] = tg; mx[x] = mi[x] = tg; s[x] = tg * (r-l+1); } inline void down(int x, int l, int r) { int mid = l+r>>1; if(hc[x]) { pushcov(ls, l, mid, cov[x]); pushcov(rs, mid+1, r, cov[x]); cov[x] = hc[x] = 0; } if(tag[x]) { pushtag(ls, l, mid, tag[x]); pushtag(rs, mid+1, r, tag[x]); tag[x] = 0; } } inline void build(int x, int l, int r) { tag[x] = cov[x] = hc[x] = 0; if(l == r) { mx[x] = mi[x] = s[x] = a[l]; return ; } int mid = l+r>>1; build(ls, l, mid); build(rs, mid+1, r); up(x); } inline void edt(int x, int l, int r, int L, int R, ll p) { if(L <= l && r <= R) { pushtag(x, l, r, p); return ; } down(x, l, r); int mid = l+r>>1; if(L <= mid) edt(ls, l, mid, L, R, p); if(R > mid) edt(rs, mid+1, r, L, R, p); up(x); } inline void cover(int x, int l, int r, int L, int R, ll p) { if(L <= l && r <= R) { pushcov(x, l, r, p); return ; } down(x, l, r); int mid = l+r>>1; if(L <= mid) cover(ls, l, mid, L, R, p); if(R > mid) cover(rs, mid+1, r, L, R, p); up(x); } inline ll sum(int x, int l, int r, int L, int R) { if(L <= l && r <= R) return s[x]; down(x, l, r); int mid = l+r>>1; ll ret = 0; if(L <= mid) ret += sum(ls, l, mid, L, R); if(R > mid) ret += sum(rs, mid+1, r, L, R); return ret; } inline ll gmax(int x, int l, int r, int L, int R) { if(L <= l && r <= R) return mx[x]; down(x, l, r); int mid = l+r>>1; if(R <= mid) return gmax(ls, l, mid, L, R); else if(L > mid) return gmax(rs, mid+1, r, L, R); else return max(gmax(ls, l, mid, L, R), gmax(rs, mid+1, r, L, R)); } inline ll gmin(int x, int l, int r, int L, int R) { if(L <= l && r <= R) return mi[x]; down(x, l, r); int mid = l+r>>1; if(R <= mid) return gmin(ls, l, mid, L, R); else if(L > mid) return gmin(rs, mid+1, r, L, R); else return min(gmin(ls, l, mid, L, R), gmin(rs, mid+1, r, L, R)); } }T; int main() { register int Q, x, y, z; register char op[23]; cin >> n >> Q; for (int i=1; i<=n; ++i) scanf(LLD, a+i); T.build(1, 1, n); while(Q--) { scanf("%s%d%d", op, &x, &y); if(op[1] == ‘d‘) { scanf(LLD, &z); T.edt(1, 1, n, x, y, z); } else if(op[1] == ‘e‘) { scanf(LLD, &z); T.cover(1, 1, n, x, y, z); } else if(op[1] == ‘u‘) printf(LLD "\n", T.sum(1, 1, n, x, y)); else if(op[1] == ‘a‘) printf(LLD "\n", T.gmax(1, 1, n, x, y)); else printf(LLD "\n", T.gmin(1, 1, n, x, y)); } return 0; }
标签：max clu 节点 scan bzoj uniq 区间 stat 需要
原文地址：http://www.cnblogs.com/galaxies/p/template.html
